Problem: Simplify and expand the following expression: $ \dfrac{1}{q - 9}+ \dfrac{3}{q + 1}+ \dfrac{5q}{q^2 - 8q - 9} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{5q}{q^2 - 8q - 9} = \dfrac{5q}{(q - 9)(q + 1)}$ Now we have: $ \dfrac{1}{q - 9}+ \dfrac{3}{q + 1}+ \dfrac{5q}{(q - 9)(q + 1)} $ The least common multiple of the denominators is: $ (q - 9)(q + 1)$ In order to get the first term over $(q - 9)(q + 1)$ , multiply by $\dfrac{q + 1}{q + 1}$ $ \dfrac{1}{q - 9} \times \dfrac{q + 1}{q + 1} = \dfrac{q + 1}{(q - 9)(q + 1)} $ In order to get the second term over $(q - 9)(q + 1)$ , multiply by $\dfrac{q - 9}{q - 9}$ $ \dfrac{3}{q + 1} \times \dfrac{q - 9}{q - 9} = \dfrac{3(q - 9)}{(q - 9)(q + 1)} $ Now we have: $ \dfrac{q + 1}{(q - 9)(q + 1)} + \dfrac{3(q - 9)}{(q - 9)(q + 1)} + \dfrac{5q}{(q - 9)(q + 1)} $ $ = \dfrac{ q + 1 + 3(q - 9) + 5q} {(q - 9)(q + 1)} $ Expand: $ = \dfrac{q + 1 + 3q - 27 + 5q}{q^2 - 8q - 9} $ $ = \dfrac{9q - 26}{q^2 - 8q - 9}$